Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int a = sc.nextInt();
		int b = sc.nextInt();
		long[] v = new long[n];
		for(int i = 0; i < n; i++){
		    v[i] = sc.nextLong();
		}
		Arrays.sort(v);
		
		// 計算
		long result = 0;
		long sum = 0;
		double ave = 0;
		int min = 0;
		int max = 0;
		int all = 0;
		for(int i = 0; i < n; i++){
		    if(i+1 <= a){
		        sum += v[n-1-i];
		        ave = (double)sum/(i+1);
		        if(i > 0 && v[n-i] == v[n-1-i]){
		            min++;
		            max++;
		            all++;
		        }else{
		            min = 1;
		            max = 1;
		            all = 1;
		        }
		    }else if(i+1 <= b && ave < v[n-1-i]){
		        sum += v[n-1-i];
		        ave = (double)sum/(i+1);
		        if(i > 0 && v[n-i] == v[n-1-i]){
		            min++;
		            max++;
		            all++;
		        }else{
		            min = 1;
		            max = 1;
		            all = 1;
		        }
		    }else if(i+1 <= b && ave == v[n-1-i]){
		        if(i > 0 && v[n-i] == v[n-1-i]){
		            max++;
		            all++;
		        }else{
		            break;
		        }
		    }else if(v[n-i] == v[n-1-i]){
		        all++;
		    }
		}
		
		for(int i = min; i <= max; i++){
		    result += combination(all, i);
		}
		
		// 出力
		System.out.println(ave);
		System.out.println(result);
	}
	
	public static long combination(long n, long k) {
	    if (n < k || n < 0 || k < 0) return 0;
	    long res = 1;
	    if(k > n-k) k = n-k;
	    for(int i = 1; i <= k; i++){
	        res *= n-k+i;
	        res /= i;
	    }
		return res;
	}
	
	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
}